Question

Two bulbs A and B are rated as 90W-120V and 60W-120V respectively. They are connected in parallel across 120V source. Find the current and resistance. Which had consumed more energy?

Answer 1

First Bulb Rated = 90W -120V

Second Bulb Rated= 60- 120V

When R =90

R=V Sqaure Upon P = (120) Sqaure Upon 90
=160


When R=60

R=V Sqaure Upon P = (120) Sqaure Upon 60
=240


R=R1 Multiply R2 Upon R1 +R2 = 160 Multiply 240 Upon 160+ 240=96 Ohm


Current (I) =V Upon R =120 Upon 96 = 1.25 Ampere (A)

Answer 2

Answer: The value of the current and the resistance in bulb A and B are 0.75 A, 0.5 A, 160 ohm and 240 ohm. Bulb A will consume more energy.

Explanation:

The expression of the power in terms of the voltage and the resistance is as follows;

$P=\frac{V^{2}}{R}$

Here, V is the voltage and R is the resistance.

Calculate the resistance for bulb A.

$P=\frac{V^{2}}{R_{1}}$

Rearrange the equation for resistance.

$R_{1}=\frac{V^{2}}{P}$

Put V= 120 V and P= 90 W to resistance for bulb A.

$R_{1}=\frac{120^{2}}{90}$

$R_{1}=160 ohm$

Calculate the resistance for bulb B.

$P=\frac{V^{2}}{R_{2}}$

Rearrange the equation for resistance.

$R_{2}=\frac{V^{2}}{P}$

Put V= 120 V and P= 60 W to resistance for bulb A.

$R_{2}=\frac{120^{2}}{60}$

$R_{2}=240 ohm$

The expression for the ohm's law is as follows;

V=IR

Calculate the current in the bulb A.

Put R=160 ohm and V=120 V.

$I=\frac{120}{160}$

$I=0.75 A$

Calculate the current in the bulb B.

Put R=240 ohm and V=120 V.

$I=\frac{120}{240}$

$I=0.5 A$

The expression for the energy in terms of power and time is as follows;

E=Pt

Here, P is the power, E is the energy and t is the time taken.

The power of bulb A is more than the bulb B.

Therefore, bulb A will consume more energy.